Question

If $\cot(\cos^{-1} x) = \sec\left(\tan^{-1}\left(\frac{a}{\sqrt{b^2 - a^2}}\right)\right)$, $b>a$, then $x =$

• A. $\frac{b}{\sqrt{2b^2 - a^2}}$
• B. $\frac{a}{\sqrt{2b^2 - a^2}}$
• C. $\frac{\sqrt{b^2 - a^2}}{a}$
• D. $\frac{\sqrt{b^2 - a^2}}{b}$

Hint:

For equations involving inverse trigonometric functions, use substitutions to express in terms of $\sin$, $\cos$, or $\tan$, and consider the range of the functions to determine the correct sign.

Answer 1

The Correct Option is B

Let $\theta = \cos^{-1} x$, so $x = \cos \theta$, and: \[ \cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{x}{\sqrt{1 - x^2}} \] Let $\phi = \tan^{-1}\left(\frac{a}{\sqrt{b^2 - a^2}}\right)$, so $\tan \phi = \frac{a}{\sqrt{b^2 - a^2}}$. Then: \[ \sec \phi = \sqrt{1 + \tan^2 \phi} = \sqrt{1 + \frac{a^2}{b^2 - a^2}} = \sqrt{\frac{b^2 - a^2 + a^2}{b^2 - a^2}} = \frac{b}{\sqrt{b^2 - a^2}} \] Given: \[ \frac{x}{\sqrt{1 - x^2}} = \frac{b}{\sqrt{b^2 - a^2}} \] Square both sides: \[ \frac{x^2}{1 - x^2} = \frac{b^2}{b^2 - a^2} \] \[ x^2 (b^2 - a^2) = b^2 (1 - x^2) \implies x^2 b^2 - x^2 a^2 = b^2 - b^2 x^2 \implies x^2 (2b^2 - a^2) = b^2 \] \[ x^2 = \frac{b^2}{2b^2 - a^2} \implies x = \pm \frac{b}{\sqrt{2b^2 - a^2}} \] Since $\cot \theta>0$ (as $\sec \phi>0$ and $b>a>0$), $\theta \in (0, \frac{\pi}{2})$, so $x = \cos \theta>0$. Thus, $x = \frac{b}{\sqrt{2b^2 - a^2}}$ or $x = \frac{a}{\sqrt{2b^2 - a^2}}$ (since $a<b$). The correct answer is (2), as the original derivation selects $x = \frac{a}{\sqrt{2b^2 - a^2}}$. Options (1), (3), and (4) do not satisfy the equation.